--- title: "Rational Fractions" book: "University Mathematics — Year 1" subject: math language: en chapter: 9 exercises: 12 source: https://one-course.com/books/math/3/en/chapter/9-rational-fractions --- # Chapter 9 — Rational Fractions A [rational fraction](#def-b1-fractions-field) is a quotient of [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def). The central theorem of this short chapter — partial fraction decomposition — breaks any such quotient into a sum of elementary bricks $\frac{c}{(X - a)^k}$. Beyond its algebraic interest, it is the standard machine for integrating rational functions ([Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration)) and for summing certain series ([Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series)). ## 9.1 The field $K(X)$ **Definition 9.1.** A *rational fraction* over $K$ ($= \R$ or $\C$) is a quotient $F = \frac{A}{B}$ with $A, B \in K[X]$, $B \neq 0$; two quotients $\frac AB$ and $\frac{A'}{B'}$ are identified when $AB' = A'B$. Every fraction has a *reduced form* with $\gcd(A, B) = 1$, unique up to constants. With the natural operations, the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $K(X)$ of rational fractions is a [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field). The *poles* of $F$ (in reduced form) are the roots of $B$; the *order* of a pole is its [multiplicity](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-derivative) as a root of $B$. The *degree* of $F$ is $\deg F = \deg A - \deg B \in \Z \cup \{-\infty\}$. **Example 9.2 (Reading off poles, orders and degree).** Let $F = \dfrac{X^3 - X}{X^4 - 2X^3 + X^2}$. Factor both layers: numerator $X(X-1)(X+1)$, denominator $X^2(X-1)^2$; cancel the common factor $X(X-1)$: $$ F = \frac{X + 1}{X(X - 1)} \quad\text{(reduced form).} $$ [Poles](#def-b1-fractions-field): $0$ and $1$, both *simple* — the orders are read on the reduced denominator, so the apparent double roots of the original one are irrelevant. Degree: $\deg F = 1 - 2 = -1$, visible asymptotically ($xF(x) \to 1$ as $x \to \infty$). The degree behaves like the [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) degree ($\deg FG = \deg F + \deg G$, $\deg(F + G) \leq \max$), a bookkeeping rule used constantly in the coefficient hunts below: each “limit of $xF(x)$” argument is a degree count in disguise. **Proposition 9.3 (Integer part).** Every $F = \frac AB \in K(X)$ is uniquely $F = E + \frac RB$ with $E \in K[X]$ (the *integer part*, or [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) part, of $F$) and $\deg R < \deg B$. One has $E \neq 0$ iff $\deg F \geq 0$. **Proof.** Euclidean division $A = BE + R$ ([Theorem 8.3](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-division)), divided by $B$. Uniqueness: if $E + \frac RB = E' + \frac{R'}{B}$, then $(E - E')B = R' - R$ with $\deg(R' - R) < \deg B$, forcing $E = E'$ then $R = R'$. ∎ **Example 9.4 (Reduce first, then divide).** Find the integer part of $F = \dfrac{X^3 + 1}{X^2 - 1}$. Dividing blindly: $X^3 + 1 = (X^2 - 1)X + (X + 1)$, so $F = X + \frac{X + 1}{X^2 - 1}$. But the fraction was not reduced: $X^3 + 1 = (X + 1)(X^2 - X + 1)$ and $X^2 - 1 = (X+1)(X-1)$ share the factor $X + 1$, and $$ F = \frac{X^2 - X + 1}{X - 1} = X + \frac{1}{X - 1} : $$ the same integer part $X$, but the fractional part collapses to a single brick, and the “[pole](#def-b1-fractions-field)” at $-1$ was never a [pole](#def-b1-fractions-field) at all. Always reduce to lowest terms before hunting [poles](#def-b1-fractions-field): the [poles](#def-b1-fractions-field) of $F$ are the roots of the *reduced* denominator. (The integer part is insensitive to the simplification, as the uniqueness in [Proposition 9.3](#prop-b1-fractions-integerpart) guarantees.) ## 9.2 Partial fraction decomposition over $\C$ **Theorem 9.5 (Decomposition over C\CC).** Let $F = \frac AB \in \C(X)$ be in reduced form, with $B = c\,(X - a_1)^{m_1} \cdots (X - a_r)^{m_r}$. Then $F$ is, in a unique way, $$ F = E + \sum_{i=1}^{r} \sum_{k=1}^{m_i} \frac{c_{i,k}}{(X - a_i)^k}, \qquad E \in \C[X],\ c_{i,k} \in \C , $$ $E$ being the integer part of $F$. **Proof.** By [Proposition 9.3](#prop-b1-fractions-integerpart) we may assume $\deg A < \deg B$ and prove the sum decomposition with $E = 0$. *Splitting the [poles](#def-b1-fractions-field).* Write $B = (X - a_1)^{m_1} B_1$ with $B_1(a_1) \neq 0$. The [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) $(X-a_1)^{m_1}$ and $B_1$ are [coprime](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#cor-b1-arith-bezout) (no common root), so by Bézout in $\C[X]$ (see the remark in [Chapter 8](https://one-course.com/books/math/3/en/chapter/8-polynomials#ch-b1-poly)) there are $U_0, V_0$ with $U_0 (X-a_1)^{m_1} + V_0 B_1 = 1$; multiplying by $A$ and setting $U = AU_0$, $V = AV_0$, then dividing by $B$: $$ \frac AB = \frac{V}{(X - a_1)^{m_1}} + \frac{U}{B_1} . $$ The degree conditions can be enforced: divide $V$ by $(X-a_1)^{m_1}$, say $V = (X-a_1)^{m_1}Q + V_1$ with $\deg V_1 < m_1$, and absorb the quotient into the second term ($U_1 = U + QB_1$): $$ \frac AB = \frac{V_1}{(X - a_1)^{m_1}} + \frac{U_1}{B_1}, \qquad \deg V_1 < m_1 ; $$ comparing degrees ($\deg A < \deg B$ and $\deg V_1 < m_1$) forces $\deg U_1 < \deg B_1$ as well. Iterating on $\frac{U_1}{B_1}$, [pole](#def-b1-fractions-field) after [pole](#def-b1-fractions-field), reduces everything to the [one-pole](#def-b1-fractions-field) case below. *One [pole](#def-b1-fractions-field).* For $\frac{A}{(X-a)^m}$ with $\deg A < m$: expand $A$ in powers of $(X - a)$, $A = \sum_{j=0}^{m-1} \alpha_j (X - a)^j$ (Taylor expansion of a [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def), as in the proof of [Proposition 8.11](https://one-course.com/books/math/3/en/chapter/8-polynomials#prop-b1-poly-multiplicity)); dividing gives exactly the bricks $\frac{\alpha_j}{(X-a)^{m-j}}$. Concretely, for $\frac{X^2 + 1}{(X+2)^3}$: substituting $X = Y - 2$, $$ X^2 + 1 = (Y - 2)^2 + 1 = Y^2 - 4Y + 5 , \qquad\text{so}\qquad \frac{X^2+1}{(X+2)^3} = \frac1{Y} - \frac4{Y^2} + \frac5{Y^3} $$ with $Y = X + 2$: the three bricks appear by mere division of the shifted expansion — the fastest route whenever a single high-order [pole](#def-b1-fractions-field) is involved, and the recommended one for [Exercise 9.3](#exo-b1-fractions-3). *Uniqueness.* Suppose two decompositions coincide; their difference is an identity $0 = \sum_{i,k} \frac{d_{i,k}}{(X - a_i)^k}$. Multiply through by $(X - a_1)^{m_1}$: every term acquires a factor vanishing at $a_1$ *except* the one with $i = 1$, $k = m_1$, whose coefficient becomes $d_{1,m_1}$ plus terms carrying at least one factor $(X - a_1)$. Evaluating at $a_1$ (legitimate: after the multiplication, no [pole](#def-b1-fractions-field) remains at $a_1$) gives $d_{1,m_1} = 0$. With the top coefficient gone, repeat with $(X - a_1)^{m_1 - 1}$, and so on downward to $k = 1$; then move to the next [pole](#def-b1-fractions-field). All $d_{i,k}$ vanish: the decomposition is unique. ∎ **Method 9.6 (Computing the coefficients).** In practice, avoid Bézout; combine: 1. *cover-up for the highest power*: the coefficient of $\frac{1}{(X-a)^m}$ (with $m$ the order of the [pole](#def-b1-fractions-field) $a$) is $$c_{a,m} = \Bigl[\,(X - a)^m F\,\Bigr]_{X = a};$$ for a *simple* [pole](#def-b1-fractions-field) of $F = \frac AB$, this is $\frac{A(a)}{B'(a)}$; 2. *evaluations* at convenient points and *limits* of $xF(x)$ as $x \to \infty$ to collect linear relations for the remaining coefficients; 3. *parity* or conjugation symmetries, when present, to cut the work in half. **Remark 9.7 (Common pitfalls with partial fractions).** 1. *Skipping the integer part.* The brick decomposition applies to fractions of degree $< 0$ ; when $\deg A \geq \deg B$ , divide first ( [Proposition 9.3](#prop-b1-fractions-integerpart) ), or the coefficient hunt will produce contradictions. 2. *Cover-up beyond its scope.* Multiplying by $(X - a)^k$ and evaluating at $a$ yields the coefficient only for $k = m$ , the *full* order of the [pole](#def-b1-fractions-field) ; the lower-order coefficients require other relations (limits, evaluations) — see [Example 9.9](#ex-b1-fractions-multiple) . 3. *Forgetting to reduce.* [Poles](#def-b1-fractions-field) are read on the *reduced* form; a common factor between numerator and denominator creates phantom [poles](#def-b1-fractions-field) ( [Example 9.4](#ex-b1-fractions-integerpart) ). 4. *Wrong brick shapes over $\R$.* Above an irreducible quadratic the numerators are *affine* ( $\alpha X + \beta$ ), not constant; writing $\frac{c}{X^2 + 1}$ alone loses solutions — the correct shapes are dictated by [Theorem 9.10](#thm-b1-fractions-real) , never improvised. **Proof of the simple-pole formula.** Near a simple [pole](#def-b1-fractions-field) $a$: $B = (X - a) Q$ with $Q(a) \neq 0$, and $B' = Q + (X - a) Q'$, so $B'(a) = Q(a)$. The cover-up value is $\frac{A(a)}{Q(a)} = \frac{A(a)}{B'(a)}$. ∎ **Example 9.8.** Decompose $F = \dfrac{1}{X(X-1)(X-2)}$. Three simple [poles](#def-b1-fractions-field); cover-up at each: $$ c_0 = \frac{1}{(0-1)(0-2)} = \frac12, \quad c_1 = \frac{1}{1 \times (1 - 2)} = -1, \quad c_2 = \frac{1}{2 \times 1} = \frac12, $$ so $F = \dfrac{1/2}{X} - \dfrac{1}{X-1} + \dfrac{1/2}{X-2}$. *Check at $X = 3$:* directly, $F(3) = \frac{1}{3\cdot2\cdot1} = \frac16$; from the decomposition, $\frac{1/2}{3} - \frac{1}{2} + \frac{1/2}{1} = \frac16 - \frac12 + \frac12 = \frac16$. **Example 9.9 (Multiple pole).** Decompose $F = \dfrac{X}{(X-1)^2 (X+1)}$. Shape: $\frac{a}{(X-1)^2} + \frac{b}{X-1} + \frac{c}{X+1}$. - Cover-up at the double [pole](#def-b1-fractions-field) : $a = \bigl[\frac{X}{X+1}\bigr]_{X=1} = \frac12$ . - Cover-up at $-1$ : $c = \bigl[\frac{X}{(X-1)^2}\bigr]_{X=-1} = -\frac14$ . - Limit of $xF(x)$ at $\infty$ : $0 = b + c$ , so $b = \frac14$ . $$ F = \frac{1/2}{(X-1)^2} + \frac{1/4}{X-1} - \frac{1/4}{X+1} . $$ *Check at $X = 0$:* $F(0) = 0$ and $\frac12 - \frac14 - \frac14 = 0$. ## 9.3 Decomposition over $\R$ **Theorem 9.10 (Decomposition over R\RR).** Let $F \in \R(X)$ be in reduced form with denominator $$ B = c \prod_i (X - a_i)^{m_i} \prod_j (X^2 + p_j X + q_j)^{n_j} \qquad (p_j^2 - 4q_j < 0). $$ Then $F$ decomposes uniquely as its integer part plus terms $$ \frac{c_{i,k}}{(X - a_i)^k} \quad (1 \leq k \leq m_i), \qquad \frac{\alpha_{j,l}\, X + \beta_{j,l}}{(X^2 + p_j X + q_j)^{l}} \quad (1 \leq l \leq n_j), $$ with real coefficients. **Proof.** Decompose over $\C$ ([Theorem 9.5](#thm-b1-fractions-complex)). Since $F$ is real, the coefficient over the [pole](#def-b1-fractions-field) $\conj a$ (at each order) is the conjugate of the coefficient over $a$ (apply conjugation to the decomposition and invoke uniqueness). Group each conjugate pair: $$ \frac{c}{(X - z)^l} + \frac{\conj c}{(X - \conj z)^l} = \frac{c\,(X - \conj z)^l + \conj c\,(X - z)^l}{\bigl(X^2 - 2\Re(z)X + \abs z^2\bigr)^{l}}, $$ whose numerator is its own conjugate, hence real, of degree $\leq l$; splitting off multiples of the real quadratic lowers it to degree $\leq 1$ at each level $l$ (a small downward induction). Real [poles](#def-b1-fractions-field) keep their real coefficients (conjugation fixes them). Uniqueness follows from uniqueness over $\C$. ∎ **Example 9.11 (Watching the conjugate pairing).** The mechanism of the proof, on the smallest case: over $\C$, the [poles](#def-b1-fractions-field) of $\frac1{X^2+1}$ are $\pm\iu$, with cover-up coefficients $\frac1{2\iu}$ at $\iu$ and $\frac1{-2\iu}$ at $-\iu$ — conjugates of each other, as the theorem predicts: $$ \frac{1}{X^2 + 1} = \frac{1/(2\iu)}{X - \iu} - \frac{1/(2\iu)}{X + \iu} . $$ Recombining over the common denominator: $$ \frac{1}{2\iu}\cdot\frac{(X + \iu) - (X - \iu)}{X^2 + 1} = \frac{1}{2\iu}\cdot\frac{2\iu}{X^2 + 1} = \frac{1}{X^2+1} : $$ the imaginary parts cancel and the real brick reappears intact. For real integrands one usually never leaves $\R$ — but when evaluating *sums* at complex points (as the weekend problem does with [roots of unity](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#def-b1-complex-unity)), the complex bricks are the natural currency, and this pairing is the exchange rate between the two decompositions. **Example 9.12.** Decompose $F = \dfrac{4}{(X^2+1)(X-1)^2}$ over $\R$. Shape: $\frac{aX + b}{X^2 + 1} + \frac{c}{(X-1)^2} + \frac{d}{X - 1}$. Cover-up at the double [pole](#def-b1-fractions-field): $c = \bigl[\frac{4}{X^2+1}\bigr]_{X=1} = 2$. Cover-up at the complex [pole](#def-b1-fractions-field) $\iu$ (numerator over $X^2 + 1$ evaluated via the complex decomposition, or directly): multiply by $X^2 + 1$ and set $X = \iu$: $$ a\iu + b = \frac{4}{(\iu - 1)^2} = \frac{4}{-2\iu} = 2\iu , $$ so $a = 2$, $b = 0$. Limit of $xF(x)$ at infinity: $0 = a + d$, so $d = -2$. Hence $$ F = \frac{2X}{X^2+1} + \frac{2}{(X-1)^2} - \frac{2}{X-1} . $$ *Check at $X = 0$:* $F(0) = 4$ and $0 + 2 + 2 = 4$. **Example 9.13 (A cubic denominator, start to finish).** Decompose $F = \dfrac{1}{X^3 + 1}$ over $\R$. Factor first: $X^3 + 1 = (X + 1)(X^2 - X + 1)$, the quadratic having discriminant $-3 < 0$. Shape: $\frac{a}{X+1} + \frac{bX + c}{X^2 - X + 1}$. Cover-up at the simple [pole](#def-b1-fractions-field) $-1$: $a = \bigl[\frac1{X^2 - X + 1}\bigr]_{X=-1} = \frac13$. Limit of $xF(x)$ at infinity: $0 = a + b$, so $b = -\frac13$. Evaluation at $X = 0$: $1 = a + c$, so $c = \frac23$. Hence $$ \frac{1}{X^3+1} = \frac13\Bigl(\frac{1}{X+1} + \frac{-X + 2}{X^2 - X + 1}\Bigr) , $$ confirmed at $X = 1$: left side $\frac12$, right side $\frac13\bigl(\frac12 + 1\bigr) = \frac12$. Note the economy: three unknowns, three cheap linear facts (one cover-up, one limit, one evaluation), no expansion of anything — the workflow of [Method 9.6](#met-b1-fractions-compute) in its pure form. **Remark 9.14 (What it is for).** Once decomposed, a rational function integrates term by term: bricks $\frac{1}{(x-a)^k}$ have elementary primitives, and bricks $\frac{\alpha x + \beta}{(x^2 + px + q)^l}$ reduce to $\ln$ and $\arctan$ ([Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration)). Telescoping sums are the other standard application ([Exercise 9.8](#exo-b1-fractions-8)). **Remark 9.15 (Where this chapter is used).** Partial fractions are above all a *preprocessing* step: the integration chapter ([Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration)) feeds every rational integrand through [Theorem 9.10](#thm-b1-fractions-real) before integrating, and the series chapter ([Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series)) telescopes rational terms exactly as in [Exercise 9.8](#exo-b1-fractions-8) and in the weekend problem below — which pushes the technique all the way to $\sum 1/k^2 = \pi^2/6$. The logarithmic derivative $P'/P = \sum m_i/(X - a_i)$ ([Exercise 8.11](https://one-course.com/books/math/3/en/chapter/8-polynomials#exo-b1-poly-11)) reappears whenever root locations are studied. Beyond this volume, the decomposition of $1/\chi(X)$ for a characteristic [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) $\chi$ underlies the computation of matrix powers and of Laplace transforms in the Year 2 volume: the bricks $\frac{c}{(X - a)^k}$ are the algebraic shadow of the solutions $t^{k-1}\eu^{at}$ met in [Chapter 5](https://one-course.com/books/math/3/en/chapter/5-linear-differential-equations#ch-b1-diffeq). ![The function F(x) = 1x+1 + 1x + 1x-1 of -type: strictly decreasing on each interval between its poles -1, 0, 1 (dashed). Each horizontal level > 0 is crossed exactly once per interval to the right of the first pole (marked dots): the solutions of F = interlace with the poles.](https://one-course.com/images/onecourse/chapters/math-3/b1-fractions/fig-0d4a101aa56d.svg) *The function $F(x) = \frac1{x+1} + \frac1x + \frac1{x-1}$ of [Exercise 9.12](#exo-b1-fractions-12)-type: strictly decreasing on each interval between its [poles](#def-b1-fractions-field) $-1, 0, 1$ (dashed). Each horizontal level $\lambda > 0$ is crossed exactly once per interval to the right of the first [pole](#def-b1-fractions-field) (marked dots): the solutions of $F = \lambda$ interlace with the [poles](#def-b1-fractions-field).* ## 9.4 Exercises **Exercise 9.1 ★.** Decompose over $\R$: $\dfrac{1}{X^2 - 1}$; $\;\dfrac{X}{X^2 - 3X + 2}$; $\;\dfrac{X^2 + 1}{X(X-1)}$ *(mind the integer part)*. **Solution of Exercise 9.1.** $\dfrac{1}{X^2 - 1}$: simple [poles](#def-b1-fractions-field) $\pm 1$; cover-up: $\dfrac{1/2}{X-1} - \dfrac{1/2}{X+1}$. $\dfrac{X}{X^2 - 3X + 2} = \dfrac{X}{(X-1)(X-2)}$: cover-up gives $\frac{1}{1-2} = -1$ at $1$ and $\frac{2}{2-1} = 2$ at $2$: $\dfrac{-1}{X-1} + \dfrac{2}{X-2}$. $\dfrac{X^2+1}{X(X-1)}$: the degree is $0$, so there is an integer part: dividing, $X^2 + 1 = (X^2 - X) + (X + 1)$, so $F = 1 + \frac{X+1}{X(X-1)}$. Cover-up on the remainder: $\frac{1}{-1} = -1$ at $0$, $\frac{2}{1} = 2$ at $1$: $$ F = 1 - \frac{1}{X} + \frac{2}{X-1} . $$ **Exercise 9.2 ★.** Decompose over $\R$: $\dfrac{1}{X(X^2 + 1)}$ and $\dfrac{X^3}{X^2 + X + 1}$. **Solution of Exercise 9.2.** $\dfrac{1}{X(X^2+1)}$: shape $\frac aX + \frac{bX + c}{X^2 + 1}$. Cover-up at $0$: $a = 1$. Limit of $xF$: $0 = a + b$, so $b = -1$. Evaluation at $X = 1$: $\frac12 = 1 + \frac{c - 1}{2}$, so $c = 0$: $$ \frac{1}{X(X^2+1)} = \frac 1X - \frac{X}{X^2+1} . $$ $\dfrac{X^3}{X^2+X+1}$: division: $X^3 = (X^2+X+1)(X - 1) + 1$, so $$ \frac{X^3}{X^2+X+1} = X - 1 + \frac{1}{X^2 + X + 1} , $$ already in real decomposed form (the quadratic has negative discriminant). **Exercise 9.3 ★.** Decompose $\dfrac{1}{X^2(X - 1)}$ and $\dfrac{X + 1}{(X - 1)^3}$ *(for the second, substitute $Y = X - 1$)*. **Solution of Exercise 9.3.** $\dfrac{1}{X^2(X-1)}$: shape $\frac{a}{X^2} + \frac bX + \frac{c}{X-1}$. Cover-up at the double [pole](#def-b1-fractions-field) $0$: $a = \bigl[\frac{1}{X-1}\bigr]_{0} = -1$. Cover-up at $1$: $c = 1$. Limit of $xF$: $0 = b + c$, so $b = -1$: $$ \frac{1}{X^2(X-1)} = -\frac{1}{X^2} - \frac1X + \frac{1}{X-1} . $$ $\dfrac{X+1}{(X-1)^3}$: with $Y = X - 1$, the numerator is $Y + 2$: $$ \frac{Y + 2}{Y^3} = \frac{1}{Y^2} + \frac{2}{Y^3} = \frac{1}{(X-1)^2} + \frac{2}{(X-1)^3} . $$ **Exercise 9.4 ★★.** Decompose over $\C$, then over $\R$: $\dfrac{1}{X^4 - 1}$. **Solution of Exercise 9.4.** The [poles](#def-b1-fractions-field) are the fourth [roots of unity](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#def-b1-complex-unity) $1, \iu, -1, -\iu$, all simple. [Simple-pole](#def-b1-fractions-field) formula with $B' = 4X^3$: coefficient at $a$ is $\frac{1}{4a^3} = \frac{a}{4a^4} = \frac a4$ (using $a^4 = 1$). So, over $\C$: $$ \frac{1}{X^4 - 1} = \frac{1/4}{X - 1} - \frac{1/4}{X + 1} + \frac{\iu/4}{X - \iu} - \frac{\iu/4}{X + \iu} . $$ Grouping the conjugate pair (common denominator $X^2 + 1$): $\frac{\iu}{4}\bigl(\frac{1}{X-\iu} - \frac{1}{X+\iu}\bigr) = \frac{\iu}{4}\cdot\frac{2\iu}{X^2+1} = \frac{-1/2}{X^2+1}$. Over $\R$: $$ \frac{1}{X^4 - 1} = \frac{1/4}{X-1} - \frac{1/4}{X+1} - \frac{1/2}{X^2 + 1} . $$ *Check at $X = 0$:* $-1 = -\frac14 - \frac14 - \frac12$. **Exercise 9.5 ★★.** Decompose over $\R$: $\dfrac{X^2}{(X^2 + 1)^2}$, and deduce a primitive of $x \mapsto \dfrac{x^2}{(x^2+1)^2}$ given $\int \frac{\dd x}{(x^2+1)^2} = \frac12\bigl(\arctan x + \frac{x}{x^2+1}\bigr) + C$. **Solution of Exercise 9.5.** $\dfrac{X^2}{(X^2+1)^2} = \dfrac{(X^2 + 1) - 1}{(X^2+1)^2} = \dfrac{1}{X^2+1} - \dfrac{1}{(X^2+1)^2}$. Hence a primitive: $$ \int \frac{x^2\,\dd x}{(x^2+1)^2} = \arctan x - \frac12\Bigl(\arctan x + \frac{x}{x^2+1}\Bigr) + C = \frac12\arctan x - \frac{x}{2(x^2+1)} + C . $$ **Exercise 9.6 ★★.** For $n \in \N^*$, decompose $F_n = \dfrac{n!}{X(X+1)\cdots(X+n)}$ *(simple [poles](#def-b1-fractions-field) at $0, -1, \dots, -n$; use the cover-up formula and recognize [binomial coefficients](https://one-course.com/books/math/3/en/chapter/2-counting#def-b1-counting-objects))*. **Solution of Exercise 9.6.** The [poles](#def-b1-fractions-field) $0, -1, \dots, -n$ are simple. Cover-up at $-k$: $$ c_k = \frac{n!}{\prod_{j \neq k} (-k + j)} = \frac{n!}{\bigl(\prod_{j=0}^{k-1}(j - k)\bigr) \bigl(\prod_{j=k+1}^{n}(j-k)\bigr)} = \frac{n!}{(-1)^k k!\,(n-k)!} = (-1)^k \binom nk . $$ So $$ \frac{n!}{X(X+1)\cdots(X+n)} = \sum_{k=0}^{n} \frac{(-1)^k \binom nk}{X + k} . $$ (Sanity check for $n = 1$: $\frac{1}{X(X+1)} = \frac1X - \frac{1}{X+1}$.) **Exercise 9.7 ★★.** Using the identity of [Exercise 8.11](https://one-course.com/books/math/3/en/chapter/8-polynomials#exo-b1-poly-11) for $P = X^n - 1$, prove that $$ \sum_{k=0}^{n-1} \frac{1}{X - \omega^k} = \frac{n X^{n-1}}{X^n - 1}, \qquad \omega = \eu^{2\iu\pi/n}, $$ and evaluate both sides at $X = 2$ for $n = 4$ as a check. **Solution of Exercise 9.7.** $P = X^n - 1$ has the $n$ simple roots $\omega^k$ ([Theorem 3.14](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#thm-b1-complex-roots)), so the logarithmic-derivative identity of [Exercise 8.11](https://one-course.com/books/math/3/en/chapter/8-polynomials#exo-b1-poly-11) reads $$ \sum_{k=0}^{n-1} \frac{1}{X - \omega^k} = \frac{P'(X)}{P(X)} = \frac{n X^{n-1}}{X^n - 1} . $$ At $X = 2$, $n = 4$: right side $= \frac{4 \times 8}{15} = \frac{32}{15}$. Left side: $\frac{1}{2-1} + \frac{1}{2+1} + \frac{1}{2 - \iu} + \frac{1}{2 + \iu} = 1 + \frac13 + \frac{4}{5} = \frac{15 + 5 + 12}{15} = \frac{32}{15}$, using $\frac{1}{2-\iu} + \frac{1}{2+\iu} = \frac{4}{5}$. **Exercise 9.8 ★★.** Decompose $\dfrac{1}{k(k+1)(k+2)}$ and compute $$ S_n = \sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)}, \qquad\text{then}\qquad \lim_{n \to \infty} S_n . $$ **Solution of Exercise 9.8.** Cover-up: $\dfrac{1}{k(k+1)(k+2)} = \dfrac{1/2}{k} - \dfrac{1}{k+1} + \dfrac{1/2}{k+2}$. Rewrite as a telescoping difference: $$ \frac{1}{k(k+1)(k+2)} = \frac12\Bigl(\frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)}\Bigr), $$ (expand to check — or subtract the two decompositions). Summing: $$ S_n = \frac12\Bigl(\frac{1}{1 \times 2} - \frac{1}{(n+1)(n+2)}\Bigr) = \frac14 - \frac{1}{2(n+1)(n+2)} \xrightarrow[n \to \infty]{} \frac14 . $$ **Exercise 9.9 ★★★.** Let $P \in \R[X]$ be [monic](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) of degree $n$ with $n$ distinct real roots $x_1 < \dots < x_n$. Prove that $$ \sum_{i=1}^{n} \frac{1}{P'(x_i)} = 0 \quad (n \geq 2), \qquad \sum_{i=1}^{n} \frac{x_i^{\,n-1}}{P'(x_i)} = 1 . $$ *Hint: decompose $\frac{X^m}{P}$ for $m \leq n - 1$ and look at the coefficient decay at infinity — or use [Lagrange interpolation](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-lagrange) ([Theorem 8.23](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-lagrange)) of the monomial $X^m$ at the nodes $x_i$.* **Solution of Exercise 9.9.** Decompose, for $0 \leq m \leq n - 1$, the fraction $\frac{X^m}{P}$ (degree $m - n \leq -1$, [poles](#def-b1-fractions-field) simple): the [simple-pole](#def-b1-fractions-field) formula gives $$ \frac{X^m}{P} = \sum_{i=1}^{n} \frac{x_i^m / P'(x_i)}{X - x_i} . $$ Multiply by $X$ and let $X \to +\infty$: the left side tends to the limit of $X^{m+1}/P$, which is $0$ if $m \leq n - 2$ and $1$ if $m = n - 1$ ($P$ is [monic](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) of degree $n$); the right side tends to $\sum_i \frac{x_i^m}{P'(x_i)}$. Hence $$ \sum_{i=1}^{n} \frac{x_i^{m}}{P'(x_i)} = \begin{cases} 0 & \text{for } 0 \leq m \leq n-2,\\ 1 & \text{for } m = n - 1, \end{cases} $$ which contains both announced identities ($m = 0$ requires $n \geq 2$). (Interpretation via [Theorem 8.23](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-lagrange): these sums are the leading coefficients of the Lagrange interpolants of $X^m$, and interpolating a [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) of degree $\leq n-1$ at $n$ points reproduces it exactly.) **Exercise 9.10 ★★.** Decompose $\dfrac{1}{X(X+1)^2}$ over $\R$. Admitting the value $\sum_{k \geq 1} \frac1{k^2} = \frac{\pi^2}6$ (proved in this chapter’s weekend problem), deduce $$ \sum_{n=1}^{\infty} \frac{1}{n(n+1)^2} = 2 - \frac{\pi^2}{6} . $$ **Solution of Exercise 9.10.** Shape $\frac aX + \frac b{X+1} + \frac c{(X+1)^2}$. Cover-up at $0$: $a = 1$; cover-up at the double [pole](#def-b1-fractions-field): $c = \bigl[\frac1X \bigr]_{X=-1} = -1$; limit of $xF(x)$ at infinity: $0 = a + b$, so $b = -1$: $$ \frac{1}{X(X+1)^2} = \frac1X - \frac1{X+1} - \frac1{(X+1)^2} . $$ Summing for $n = 1, \dots, N$: the first two bricks telescope to $1 - \frac1{N+1}$, and the third contributes $-\sum_{k=2}^{N+1} \frac1{k^2}$. Letting $N \to \infty$ and using $\sum_{k\geq1} \frac1{k^2} = \frac{\pi^2}6$: $$ \sum_{n=1}^{\infty}\frac{1}{n(n+1)^2} = 1 - \Bigl(\frac{\pi^2}6 - 1\Bigr) = 2 - \frac{\pi^2}6 \approx 0.355 . $$ **Exercise 9.11 ★★.** Decompose over $\R$: $\dfrac{1}{(X^2+1)(X^2+4)}$, then $\dfrac{X^2}{(X^2+1)(X^2+4)}$. *Hint: both denominators are [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) in $X^2$: decompose $\frac1{(Y+1)(Y+4)}$ first.* **Solution of Exercise 9.11.** In the variable $Y$: $\frac1{(Y+1)(Y+4)} = \frac{1/3}{Y+1} - \frac{1/3}{Y+4}$ (cover-up at $-1$ and $-4$), so $$ \frac{1}{(X^2+1)(X^2+4)} = \frac13\,\frac1{X^2+1} - \frac13\,\frac1{X^2+4} . $$ Likewise $\frac{Y}{(Y+1)(Y+4)} = \frac{-1/3}{Y+1} + \frac{4/3}{Y+4}$, so $$ \frac{X^2}{(X^2+1)(X^2+4)} = -\frac13\,\frac1{X^2+1} + \frac43\,\frac1{X^2+4} . $$ (Check at $X = 0$: $0 = \frac13(-1 + 1)$.) These are already the real decompositions: the numerators over the irreducible quadratics happen to be constants. **Exercise 9.12 ★★★.** (Secular equations) Let $F = \sum_{i=1}^{r} \frac{c_i}{X - p_i}$ with $p_1 < p_2 < \dots < p_r$ real and all $c_i > 0$. 1. Show that $F$ is strictly decreasing on each interval of its domain, and give its limits at $\pm\infty$ and on both sides of each [pole](#def-b1-fractions-field) . 2. Deduce that for every $\lambda > 0$ , the equation $F(x) = \lambda$ has exactly $r$ real solutions, one in each interval $\intoo{p_i}{p_{i+1}}$ and one beyond $p_r$ . (Such equations govern eigenvalue perturbations; the intermediate value theorem is used at High School level here and proved in [Chapter 13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ch-b1-continuity) .) **Solution of Exercise 9.12.** 1. On each interval avoiding the [poles](#def-b1-fractions-field) , $F'(x) = -\sum_i \frac{c_i}{(x - p_i)^2} < 0$ : strictly decreasing. As $x \to \pm\infty$ , every brick tends to $0$ : $F \to 0$ , from above at $+\infty$ (all bricks positive there) and from below at $-\infty$ . As $x \to p_i^+$ , the brick $\frac{c_i}{x - p_i}$ blows up to $+\infty$ and the others stay bounded: $F \to +\infty$ ; likewise $F \to -\infty$ as $x \to p_i^-$ . 2. Fix $\lambda > 0$ . On $\intoo{-\infty}{p_1}$ : $F$ decreases from $0^-$ to $-\infty$ , so $F < 0 < \lambda$ : no solution. On each $\intoo{p_i}{p_{i+1}}$ ( $1 \leq i \leq r-1$ ): $F$ decreases from $+\infty$ to $-\infty$ , hence takes the value $\lambda$ exactly once (intermediate value property plus strict monotonicity). On $\intoo{p_r}{+\infty}$ : $F$ decreases from $+\infty$ to $0^+$ , again exactly one solution. Total: exactly $r$ solutions, interlaced with the [poles](#def-b1-fractions-field) . ## 9.5 Problem: Partial fractions as an engine **Problem 9.1.** Partial fraction decomposition looks like bookkeeping; this problem shows it is an engine. Fed with the fraction $\frac1{X(X+1)\cdots(X+k)}$, it telescopes whole families of sums in closed form; fed with $\frac1{X^n - 1}$, it produces trigonometric identities such as $$ \sum_{k=1}^{n-1}\frac{1}{\sin^2\frac{k\pi}{n}} = \frac{n^2-1}{3} ; $$ and, pushed one step further, that identity squeezes out one of the most celebrated formulas in mathematics, Euler’s $$ \sum_{k=1}^{\infty}\frac1{k^2} = \frac{\pi^2}{6} $$ — here obtained with nothing beyond this chapter’s algebra and high-school trigonometry. Throughout, $\omega = \eu^{2\iu\pi/n}$; limits of sequences are used at High School level ([Chapter 11](https://one-course.com/books/math/3/en/chapter/11-sequences#ch-b1-seq) formalizes them). **Part I — The telescope.** 1. Decompose $\frac1{X(X+1)}$ and compute $\sum_{n=1}^{N} \frac1{n(n+1)}$ exactly; conclude that the sum tends to $1$ . 2. Same for $\frac1{X(X+2)}$ : show $\sum_{n=1}^{N} \frac1{n(n+2)} = \frac12\bigl(\frac32 - \frac1{N+1} - \frac1{N+2}\bigr) \to \frac34$ . (With a gap, *two* boundary terms survive at each end.) 3. Formalize the mechanism: if $F(X) = G(X) - G(X+1)$ for some rational $G$ without [poles](#def-b1-fractions-field) in $\intco1{+\infty}$ , then $\sum_{n=1}^N F(n) = G(1) - G(N+1)$ . Recover the value $\frac14$ of [Exercise 9.8](#exo-b1-fractions-8) by exhibiting the witness $G$ for $F = \frac1{X(X+1)(X+2)}$ . 4. Prove the general factorial telescope: for $k \geq 1$, $$\frac{1}{X(X+1)\cdots(X+k)} = \frac1k\biggl(\frac{1}{X(X+1)\cdots(X+k-1)} - \frac{1}{(X+1)\cdots(X+k)}\biggr),$$ and deduce $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+k)} = \frac{1}{k \cdot k!} .$$ Check the case $k = 2$ against question 3. 5. Evaluate the decomposition of [Exercise 9.6](#exo-b1-fractions-6) at well-chosen points to prove $$\sum_{j=0}^{n}(-1)^j\binom nj\,\frac1{j+1} = \frac1{n+1}, \qquad \sum_{j=0}^{n}(-1)^j\binom nj\,\frac1{j+2} = \frac1{(n+1)(n+2)} .$$ **Part II — The fraction $1/(X^n - 1)$.** 6. Show by the cover-up formula that $$\frac{1}{X^n - 1} = \frac1n\sum_{k=0}^{n-1}\frac{\omega^k}{X - \omega^k} .$$ 7. Two sanity checks: verify the formula directly for $n = 2$ , and show that the sum of the $n$ coefficients vanishes for $n \geq 2$ — explain why it must (consider $xF(x)$ as $x \to \infty$ ). 8. Re-derive by cover-up the identity of [Exercise 9.7](#exo-b1-fractions-7) : $\frac{nX^{n-1}}{X^n-1} = \sum_k \frac1{X - \omega^k}$ . 9. Group conjugate [poles](#def-b1-fractions-field) to prove the real decomposition: with $\theta_k = \frac{2k\pi}n$, $$\frac{\omega^k}{X - \omega^k} + \frac{\omega^{n-k}}{X - \omega^{n-k}} = \frac{2\cos\theta_k\,X - 2} {X^2 - 2\cos\theta_k\,X + 1} ,$$ and write down the full real decomposition of $\frac1{X^n-1}$ (distinguish $n$ odd and $n$ even). 10. Specialize to $n = 4$ and check against [Exercise 9.4](#exo-b1-fractions-4) . **Part III — Trigonometric sums, and Euler’s $\pi^2/6$.** Let $P = 1 + X + \dots + X^{n-1}$, whose roots are $\omega, \omega^2, \dots, \omega^{n-1}$ (all simple). 11. Using $\frac{P'}P = \sum_{k=1}^{n-1}\frac1{X - \omega^k}$ ([Exercise 8.11](https://one-course.com/books/math/3/en/chapter/8-polynomials#exo-b1-poly-11)), prove $$\sum_{k=1}^{n-1}\frac{1}{1 - \omega^k} = \frac{n-1}2 .$$ 12. Prove $\dfrac1{1 - \eu^{\iu\theta}} = \dfrac12 + \dfrac\iu2\,\frac{\cos(\theta/2)}{\sin(\theta/2)}$ for $\theta \notin 2\pi\Z$ (half-angle factorization, [Method 3.11](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#met-b1-complex-trig) ), and deduce from question 11 that $\sum_{k=1}^{n-1} \frac{\cos(k\pi/n)}{\sin(k\pi/n)} = 0$ — also visible by the symmetry $k \leftrightarrow n - k$ . 13. Differentiating the identity of question 11 (i.e. using $\bigl(\frac{P'}P\bigr)' = \frac{P''}P - \bigl(\frac{P'}P\bigr)^2$ evaluated at $X = 1$), prove $$\sum_{k=1}^{n-1}\frac{1}{(1 - \omega^k)^2} = \frac{(n-1)(5-n)}{12} .$$ 14. Writing $\cot t = \frac{\cos t}{\sin t}$, deduce from questions 12–13 the two closed forms $$\sum_{k=1}^{n-1}\cot^2\frac{k\pi}n = \frac{(n-1)(n-2)}3, \qquad \sum_{k=1}^{n-1}\frac1{\sin^2\frac{k\pi}n} = \frac{n^2-1}3 .$$ 15. Verify both formulas by hand for $n = 3$ and $n = 4$ . 16. Prove the inequalities $\cot t < \frac1t < \frac1{\sin t}$ for $t \in \intoo0{\frac\pi2}$ (from $\sin t < t < \tan t$), and deduce, for $n = 2m + 1$ and $1 \leq k \leq m$: $$\cot^2\frac{k\pi}n \;<\; \frac{n^2}{k^2\pi^2} \;<\; \frac1{\sin^2\frac{k\pi}n} .$$ 17. Sum these inequalities for $k = 1, \dots, m$ (using the symmetry $k \leftrightarrow n - k$ to halve the formulas of question 14) and squeeze: $$\sum_{k=1}^{\infty}\frac1{k^2} = \frac{\pi^2}6 .$$ **Part IV — Higher bricks.** 18. By squaring the decomposition of $\frac1{X^2-1}$ and re-decomposing the cross term, prove $$\frac1{(X^2-1)^2} = \frac14\Bigl(\frac1{(X-1)^2} + \frac1{(X+1)^2}\Bigr) - \frac14\Bigl(\frac1{X-1} - \frac1{X+1}\Bigr),$$ and check it at $X = 0$. 19. Combine question 18, the telescope, and Euler’s value (question 17) to prove $$\sum_{n=2}^{\infty}\frac1{(n^2-1)^2} = \frac{\pi^2}{12} - \frac{11}{16} ,$$ and confirm the value numerically to three decimals. 20. Differentiate the identity of question 8 to obtain a closed form for $\sum_{k=0}^{n-1}\frac1{(X - \omega^k)^2}$ , and check it at $X = 2$ , $n = 2$ . 21. Prove that for $k \geq 2$, $$\sum_{n=k}^{\infty}\frac{1}{\binom nk} = \frac{k}{k-1} .$$ *(Reduce to question 4 by writing $1/\binom nk$ with factorials.)* 22. For $k = 3$ , give the exact partial sum $\sum_{n=3}^{N}\frac1{\binom n3}$ and its limit. **Part V — Synthesis.** 23. As a capstone computation, write out the complete real decomposition of $\dfrac1{X^6 - 1}$ and check it at $X = 0$ . 24. Where exactly did the problem use: (i) the uniqueness of the decomposition; (ii) the [roots of unity](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#def-b1-complex-unity) from [Chapter 3](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#ch-b1-complex) ; (iii) the logarithmic derivative from [Exercise 8.11](https://one-course.com/books/math/3/en/chapter/8-polynomials#exo-b1-poly-11) ? One sentence each. 25. Synthesis, in a short paragraph: one algebraic identity — breaking a fraction into bricks — generated exact sums, trigonometric identities, and $\pi^2/6$ . Comment on the division of labor between algebra (exact decompositions, valid everywhere) and analysis (limits, squeezing), and point to where each thread is industrialized: telescoping and comparison in [Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series) , integration of the bricks in [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration) . **Solution of Problem 9.1.** **1.** Cover-up: $\frac1{X(X+1)} = \frac1X - \frac1{X+1}$. The sum telescopes: $$ \sum_{n=1}^{N}\Bigl(\frac1n - \frac1{n+1}\Bigr) = 1 - \frac1{N+1} \longrightarrow 1 . $$ **2.** $\frac1{X(X+2)} = \frac{1/2}X - \frac{1/2}{X+2}$. Summing, the terms $\frac1n$ survive for $n = 1, 2$ and the terms $-\frac1{n+2}$ survive for $n = N-1, N$: $$ \sum_{n=1}^{N}\frac1{n(n+2)} = \frac12\Bigl(1 + \frac12 - \frac1{N+1} - \frac1{N+2}\Bigr) \longrightarrow \frac34 . $$ **3.** If $F(X) = G(X) - G(X+1)$, then $\sum_{n=1}^N F(n) = \sum_{n=1}^N\bigl(G(n) - G(n+1)\bigr) = G(1) - G(N+1)$: all intermediate values cancel in pairs. For $F = \frac1{X(X+1)(X+2)}$, the witness is $G(X) = \frac1{2X(X+1)}$: $$ G(X) - G(X+1) = \frac{(X+2) - X}{2X(X+1)(X+2)} = F(X) , $$ so $\sum_{n=1}^N F(n) = \frac14 - \frac1{2(N+1)(N+2)} \to \frac14$, the value of [Exercise 9.8](#exo-b1-fractions-8). **4.** Put the right-hand side over the common denominator $X(X+1)\cdots(X+k)$: $$ \frac1k\cdot\frac{(X + k) - X}{X(X+1)\cdots(X+k)} = \frac{1}{X(X+1)\cdots(X+k)} , $$ which is the identity. So $F_k(X) = \frac1k\bigl(G_k(X) - G_k(X+1)\bigr)$ with $G_k(X) = \frac1{X(X+1)\cdots(X+k-1)}$, and question 3’s mechanism gives $$ \sum_{n=1}^{N}\frac1{n(n+1)\cdots(n+k)} = \frac1k\Bigl(\frac1{k!} - G_k(N+1)\Bigr) \longrightarrow \frac1{k\cdot k!} , $$ since $G_k(1) = \frac1{k!}$ and $G_k(N+1) \to 0$. For $k = 2$: $\frac1{2\cdot2} = \frac14$, matching question 3. **5.** [Exercise 9.6](#exo-b1-fractions-6) gives $\frac{n!}{X(X+1)\cdots(X+n)} = \sum_{j=0}^n \frac{(-1)^j\binom nj}{X + j}$. Evaluate at $X = 1$: the left side is $\frac{n!}{(n+1)!} = \frac1{n+1}$, the right side is $\sum_j(-1)^j\binom nj\frac1{1+j}$: first identity. At $X = 2$: the left side is $\frac{n!}{2\cdot3\cdots(n+2)} = \frac{n!\cdot1}{(n+2)!} = \frac1{(n+1)(n+2)}$, the right side $\sum_j(-1)^j\binom nj\frac1{2+j}$: second identity. **6.** The [poles](#def-b1-fractions-field) $\omega^k$ are simple, and the [simple-pole](#def-b1-fractions-field) formula of [Method 9.6](#met-b1-fractions-compute) gives the coefficient $$ \frac{1}{\bigl(nX^{n-1}\bigr)_{X = \omega^k}} = \frac{1}{n\,\omega^{k(n-1)}} = \frac{\omega^k}{n\,\omega^{kn}} = \frac{\omega^k}n , $$ using $\omega^{kn} = 1$. Hence $\frac1{X^n-1} = \frac1n\sum_k \frac{\omega^k}{X - \omega^k}$. **7.** For $n = 2$ ($\omega = -1$): $\frac12\bigl( \frac1{X-1} - \frac1{X+1}\bigr) = \frac12\cdot\frac{2}{X^2-1} = \frac1{X^2-1}$: correct. The coefficients sum to $\frac1n\sum_k\omega^k = 0$ for $n \geq 2$ ([Proposition 3.18](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#prop-b1-complex-sumroots)). They must: $x\,F(x) \to \sum_k c_k$ as $x \to \infty$ for any decomposition with simple [poles](#def-b1-fractions-field), while here $xF(x) = \frac{x}{x^n-1} \to 0$ since $n \geq 2$. **8.** Cover-up for $\frac{X^{n-1}}{X^n - 1}$ at $\omega^k$: $\frac{A(\omega^k)}{B'(\omega^k)} = \frac{\omega^{k(n-1)}}{n\omega^{k(n-1)}} = \frac1n$, so $\frac{X^{n-1}}{X^n-1} = \frac1n\sum_k\frac1{X - \omega^k}$ — the identity of [Exercise 9.7](#exo-b1-fractions-7) again. **9.** With $c = \omega^k$, $\conj c = \omega^{n-k}$ and $c\conj c = 1$, $c + \conj c = 2\cos\theta_k$: $$ \frac{c}{X - c} + \frac{\conj c}{X - \conj c} = \frac{c(X - \conj c) + \conj c(X - c)} {(X - c)(X - \conj c)} = \frac{2\cos\theta_k\,X - 2}{X^2 - 2\cos\theta_k\,X + 1} . $$ Grouping $k$ with $n - k$ in question 6: for odd $n$, $$ \frac1{X^n - 1} = \frac1n\Biggl(\frac1{X-1} + \sum_{k=1}^{(n-1)/2} \frac{2\cos\theta_k X - 2}{X^2 - 2\cos\theta_k X + 1}\Biggr) ; $$ for even $n$, the extra self-paired [pole](#def-b1-fractions-field) $\omega^{n/2} = -1$ contributes $\frac{-1}{X+1}$ inside the parenthesis and the pair sum runs to $\frac n2 - 1$. **10.** $n = 4$: $\theta_1 = \frac\pi2$, $\cos\theta_1 = 0$, so the pair term is $\frac{-2}{X^2+1}$ and $$ \frac1{X^4-1} = \frac14\Bigl(\frac1{X-1} - \frac1{X+1} - \frac2{X^2+1}\Bigr) , $$ the decomposition of [Exercise 9.4](#exo-b1-fractions-4). **11.** $P = \prod_{k=1}^{n-1}(X - \omega^k)$ (divide $X^n - 1$ by $X - 1$), so by [Exercise 8.11](https://one-course.com/books/math/3/en/chapter/8-polynomials#exo-b1-poly-11), $\frac{P'}P = \sum_{k=1}^{n-1}\frac1{X-\omega^k}$. At $X = 1$: $P(1) = n$ and $P'(1) = \sum_{j=1}^{n-1}j = \frac{n(n-1)}2$, whence $$ \sum_{k=1}^{n-1}\frac1{1 - \omega^k} = \frac{P'(1)}{P(1)} = \frac{n-1}2 . $$ **12.** Half-angle: $1 - \eu^{\iu\theta} = -2\iu\sin\frac\theta2\,\eu^{\iu\theta/2}$, so $$ \frac1{1 - \eu^{\iu\theta}} = \frac{\eu^{-\iu\theta/2}}{-2\iu\sin\frac\theta2} = \frac{\iu\cos\frac\theta2 + \sin\frac\theta2} {2\sin\frac\theta2} = \frac12 + \frac\iu2\, \frac{\cos\frac\theta2}{\sin\frac\theta2} . $$ With $\theta = \theta_k = \frac{2k\pi}n$: $\frac1{1-\omega^k} = \frac12 + \frac\iu2\cot\frac{k\pi}n$. Summing over $k = 1, \dots, n-1$ and comparing with the real value $\frac{n-1}2$ of question 11: the real parts already account for everything, so $\sum_k\cot\frac{k\pi}n = 0$ — as the symmetry $\cot\frac{(n-k) \pi}n = -\cot\frac{k\pi}n$ also shows. **13.** Differentiating $\frac{P'}P = \sum_k\frac1{X - \omega^k}$: $$ \frac{P''}P - \Bigl(\frac{P'}P\Bigr)^2 = -\sum_{k=1}^{n-1}\frac1{(X - \omega^k)^2} . $$ At $X = 1$: $P''(1) = \sum_{j=2}^{n-1}j(j-1) = 2\binom n3 = \frac{n(n-1)(n-2)}3$ (hockey-stick identity, or induction), so $$ \sum_{k=1}^{n-1}\frac1{(1 - \omega^k)^2} = \Bigl(\frac{n-1}2\Bigr)^2 - \frac{(n-1)(n-2)}3 = \frac{(n-1)\bigl(3(n-1) - 4(n-2)\bigr)}{12} = \frac{(n-1)(5-n)}{12} . $$ **14.** Squaring question 12’s formula, with $c_k = \cot\frac{k\pi}n$: $$ \frac1{(1-\omega^k)^2} = \Bigl(\frac12 + \frac\iu2 c_k\Bigr)^2 = \frac14 - \frac{c_k^2}4 + \frac\iu2\,c_k . $$ Summing and using $\sum c_k = 0$ (question 12) and question 13: $\frac{n-1}4 - \frac14\sum_k c_k^2 = \frac{(n-1)(5-n)}{12}$, so $$ \sum_{k=1}^{n-1}\cot^2\frac{k\pi}n = (n-1) - \frac{(n-1)(5-n)}3 = \frac{(n-1)(n-2)}3 . $$ Then $\frac1{\sin^2t} = 1 + \cot^2t$ gives $\sum_k\frac1{\sin^2\frac{k\pi}n} = (n-1) + \frac{(n-1)(n-2)}3 = \frac{n^2-1}3$. **15.** $n = 3$: $\cot^2\frac\pi3 + \cot^2\frac{2\pi}3 = \frac13 + \frac13 = \frac23 = \frac{2\cdot1}3$; and $\frac1{\sin^2}$ sums to $\frac43 + \frac43 = \frac83 = \frac{9-1}3$. $n = 4$: $1 + 0 + 1 = 2 = \frac{3\cdot2}3$; and $2 + 1 + 2 = 5 = \frac{16-1}3$. Both formulas check. **16.** For $t \in \intoo0{\frac\pi2}$, the classical comparison $\sin t < t < \tan t$ (area or convexity argument, familiar from High School) yields, taking reciprocals, $\cot t < \frac1t < \frac1{\sin t}$, all three positive there; squaring preserves the order. With $t = \frac{k\pi}n$, $1 \leq k \leq m$, $n = 2m+1$ (so $t < \frac\pi2$): $$ \cot^2\frac{k\pi}n < \frac{n^2}{k^2\pi^2} < \frac1{\sin^2\frac{k\pi}n} . $$ **17.** By the symmetries $\cot^2\frac{(n-k)\pi}n = \cot^2\frac{k\pi}n$ and likewise for $\sin^2$, question 14’s sums halve: $\sum_{k=1}^{m}\cot^2\frac{k\pi}n = \frac{(n-1)(n-2)}6 = \frac{m(2m-1)}3$ and $\sum_{k=1}^m\frac1{\sin^2\frac{k\pi}n} = \frac{n^2-1}6 = \frac{2m(m+1)}3$. Summing question 16 over $k = 1, \dots, m$ and multiplying by $\frac{\pi^2}{n^2}$: $$ \frac{\pi^2}{(2m+1)^2}\cdot\frac{m(2m-1)}3 \;<\; \sum_{k=1}^{m}\frac1{k^2} \;<\; \frac{\pi^2}{(2m+1)^2}\cdot\frac{2m(m+1)}3 . $$ Both bounds tend to $\frac{\pi^2}6$ as $m \to \infty$ (the ratios $\frac{m(2m-1)}{(2m+1)^2}$ and $\frac{2m(m+1)}{(2m+1)^2}$ both tend to $\frac12$), so by the squeeze the increasing partial sums converge and $$ \sum_{k=1}^{\infty}\frac1{k^2} = \frac{\pi^2}6 . $$ **18.** Square $\frac1{X^2-1} = \frac12\bigl(\frac1{X-1} - \frac1{X+1}\bigr)$: $$ \frac1{(X^2-1)^2} = \frac14\Bigl(\frac1{(X-1)^2} + \frac1{(X+1)^2}\Bigr) - \frac12\cdot\frac1{(X-1)(X+1)} , $$ and re-decompose the cross term $\frac1{(X-1)(X+1)} = \frac12\bigl(\frac1{X-1} - \frac1{X+1}\bigr)$ to get the stated form. At $X = 0$: left side $1$; right side $\frac14(1 + 1) - \frac14(-1 - 1) = \frac12 + \frac12 = 1$. **19.** Sum question 18 over $n \geq 2$. With $S = \sum_{k\geq1}\frac1{k^2} = \frac{\pi^2}6$: $\sum_{n\geq2} \frac1{(n-1)^2} = S$; $\sum_{n\geq2}\frac1{(n+1)^2} = S - 1 - \frac14$; and the gap-two telescope $\sum_{n\geq2}\bigl( \frac1{n-1} - \frac1{n+1}\bigr) = 1 + \frac12 = \frac32$. Hence $$ \sum_{n=2}^{\infty}\frac1{(n^2-1)^2} = \frac14\Bigl(2S - \frac54\Bigr) - \frac14\cdot\frac32 = \frac S2 - \frac{11}{16} = \frac{\pi^2}{12} - \frac{11}{16} \approx 0.135 . $$ Numerically: $\frac19 + \frac1{64} + \frac1{225} + \frac1{576} + \frac1{1225} + \dots \approx 0.1111 + 0.0156 + 0.0044 + 0.0017 + 0.0008 + \dots \approx 0.135$: consistent. **20.** Differentiate question 8’s identity $\sum_k\frac1{X-\omega^k} = \frac{nX^{n-1}}{X^n-1}$: $$ \sum_{k=0}^{n-1}\frac1{(X - \omega^k)^2} = \frac{n^2X^{2n-2} - n(n-1)X^{n-2}(X^n - 1)}{(X^n - 1)^2} . $$ Check at $n = 2$, $X = 2$: right side $\frac{4\cdot4 - 2\cdot1\cdot3}{9} = \frac{10}9$; left side $\frac1{(2-1)^2} + \frac1{(2+1)^2} = 1 + \frac19 = \frac{10}9$. **21.** $\frac1{\binom nk} = \frac{k!\,(n-k)!}{n!} = \frac{k!}{(n-k+1)(n-k+2)\cdots n}$, a product of $k$ consecutive integers downstairs. Substituting $j = n - k + 1$ (so $j$ runs over $\N^*$ as $n$ runs from $k$): $$ \sum_{n=k}^{\infty}\frac1{\binom nk} = k!\sum_{j=1}^{\infty}\frac1{j(j+1)\cdots(j+k-1)} = \frac{k!}{(k-1)\,(k-1)!} = \frac{k}{k-1} , $$ by question 4 applied with $k - 1$ in place of $k$ (valid since $k - 1 \geq 1$). **22.** For $k = 3$: $\frac1{\binom n3} = \frac6{(n-2)(n-1)n}$, so by question 3’s partial sum (shifted), $$ \sum_{n=3}^{N}\frac1{\binom n3} = 6\sum_{j=1}^{N-2}\frac1{j(j+1)(j+2)} = 6\Bigl(\frac14 - \frac1{2(N-1)N}\Bigr) = \frac32 - \frac3{(N-1)N} \longrightarrow \frac32 , $$ the value $\frac k{k-1} = \frac32$ of question 21. **23.** $n = 6$: pairs $k = 1$ ($\theta_1 = \frac\pi3$, $2\cos\theta_1 = 1$) and $k = 2$ ($\theta_2 = \frac{2\pi}3$, $2\cos\theta_2 = -1$), plus the real [poles](#def-b1-fractions-field) $\pm1$: $$ \frac1{X^6-1} = \frac16\Bigl(\frac1{X-1} - \frac1{X+1} + \frac{X - 2}{X^2 - X + 1} + \frac{-X - 2}{X^2 + X + 1}\Bigr) . $$ At $X = 0$: left side $-1$; right side $\frac16(-1 - 1 - 2 - 2) = -1$: correct. **24.** (i) Uniqueness legitimizes every identification of coefficients — cover-up, the conjugate-pair grouping of question 9, and the differentiation tricks (questions 13, 20) all rest on it. (ii) [Roots of unity](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#def-b1-complex-unity) supplied the [poles](#def-b1-fractions-field) of $X^n - 1$, their symmetries ($k \leftrightarrow n-k$), and the half-angle algebra of question 12 ([Method 3.11](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#met-b1-complex-trig)). (iii) The logarithmic derivative $\frac{P'}P = \sum\frac{m_i}{X-a_i}$ converted information about the roots of $P = 1 + X + \dots + X^{n-1}$ into the numerical sums of questions 11 and 13 — the hinge between Parts II and III. **25.** The decomposition is a purely algebraic identity, true for every value of the variable at once; that is what makes it an engine. Substituting integers and summing turned it into telescopes (Part I); substituting [roots of unity](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#def-b1-complex-unity) and grouping conjugates turned it into trigonometric identities (Parts II and III); and only at the very last step did analysis enter — a squeeze between two closed forms — to deliver $\frac{\pi^2}6$, a [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) no finite substitution could reach. This division of labor (algebra produces exact finite identities, analysis passes to the limit) is the template for [Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series), where telescoping and comparison become systematic, and for [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration), where each brick acquires a primitive and the same decompositions compute integrals instead of sums.